---
title: "Correct horizontal distance, 28.00 m measured"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-fully-support-28-00m"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-fully-support-28-00m
site: BCT Engineering
---

# Correct horizontal distance, 28.00 m measured

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2064 Jestha, 10 marks
- Video: [Correct horizontal distance between two points - Chain Surveying -  Online Civil Engineering Notes](https://www.youtube.com/watch?v=73DrULwNlZs) — Online Civil Engineering Notes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/tape-fully-support-28-00m
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A 30 m steel tape was standardised in catenary condition at a temperature of 20°C under a pull of 5 kg and found to be 30.005 m. The tape was used to measure the distance in fully supported condition at a temperature of 25°C under a pull of 12 kg and found to be 28.00 m. The cross sectional area of tape is 0.02 cm², its weight per unit length is 22 gm/metre, Young modulus of elasticity $E = 2.0\times10^{6}$ kg/cm², coefficient of linear expansion $\alpha = 11\times10^{-6}$/°C. Find the correct horizontal distance.

## Short answer

Add sag from catenary standard to get true length 30.02678 m. Corrections: standardization +0.0250 m, temperature +0.0015 m, pull +0.0049 m. Correct distance = 28.0314 m.

## Answer

**Method.** The tape was standardised in **catenary**, so its stated length includes the effect of sag. When used **fully supported** there is no sag, so the sag in the standard must be added back to get the true tape length. $g$ is not needed (all in kg).

#### Data
- Standard: 30.005 m at $20^\circ$C and $P_0 = 5$ kg (catenary); field: $T = 25^\circ$C, $P = 12$ kg, fully supported; $L = 28.00$ m
- $A = 0.02$ cm$^2$, $E = 2.0\times 10^{6}$ kg/cm$^2$, $\alpha = 11\times 10^{-6}$/$^\circ$C, $w = 0.022$ kg/m

#### 1. Tape length
Sag at standardisation: $\dfrac{0.022^{2}\times 30^{3}}{24\times 5^{2}} = 0.02178$ m.
True tape length $= 30.005 + 0.02178 = 30.02678$ m.

$$
C_a = 28.00\times\frac{30.02678 - 30}{30} = +0.0250\ \text{m}
$$

#### 2. Temperature
$$
C_t = 11\times 10^{-6}\times(25 - 20)\times 28 = +0.0015\ \text{m}
$$

#### 3. Pull
$$
C_p = \frac{(12 - 5)\times 28}{0.02\times 2.0\times 10^{6}} = +0.0049\ \text{m}
$$

#### 4. Sag
Zero (fully supported).

#### 5. Correct horizontal distance
$$
D = 28.00 + 0.0250 + 0.0015 + 0.0049 = 28.0314\ \text{m}
$$

**Answer: correct distance $= 28.0314$ m.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The field sag correction is zero because
   A. tape is fully supported
   B. pull is high
   C. length is short
   D. Standard was in catenary
   Answer: A. No hanging, no sag. (Key point: Field sag)

2. Sag at standard 5 kg is added to 30.005 m because
   A. Field is catenary
   B. standard in catenary includes sag effect
   C. Temperature
   D. Pull
   Answer: B. True tape length differs. (Key point: Why add sag)

3. All corrections are
   A. negative
   B. mixed
   C. positive, giving a distance greater than 28.00 m
   D. zero
   Answer: C. Tape long, warm, pulled. (Key point: Final)

4. C_p uses
   A. 12 x 28/AE
   B. (12-5)/AE
   C. (12+5)
   D. (12 - 5) x 28/(A E)
   Answer: D. Difference. (Key point: Pull correction)

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